words that the arc A B is a tenth part of the circumfeference. To prove this we need only show that the angle A 0 B=36°. We have by construction, A M:M 0:: M 0:A 0, or, drawing B M and substituting A B for M 0, A M: A B :: A B: A 0. Then the triangles A M B and A O B are similar (77) having the angle A common, and the sides including it proportional. But A O B is isosceles. Therefore A M B is also isosceles, and B M=A B=O M. This makes O M B isosceles, and the angle M O BOMBO. Now B M A, being an exterior angle (49) is equal to the two opposite interior angles MOB +M B (=twice AO B. Then B A M=B MA=twice A O B, aud O BA=M A B=twice A O B. Hence all the angles of the triangle A B or 180°=five times A O B. Then A OB=one fifth of 180°=36°, and A B is the side of a regular decagon. If now, secondly, we wish to inscribe a regular . polygon of 15 sides, we have only to find one fifteenth of a circumference. For this purpose, let A C be the side of a hexagon and A B that of a decagon. Then BC will be the arc required, for B C=- of a circumference, that is one fifteenth. Lastly by joining the alternate vertices of a decagon we should have a pentagon; and by bisecting the arcs which are one fifteenth and drawing chords, we should have a polygon of 30 sides, and so on indefinitely. 94. - The circle is a regular polygon of an infinite number of sides Inscribe in the circle (fig. 63) any one of F63 the regular polygons before mentioned, for instance a hexagon, as A B C D E F. Bisect the arcs B C, C D, &c., and join the half arcs by the chords B H, HC, C Í, &c. Thus you have a regular polygon of 12 sides. Proceed in the same manner with this, and you have one of 24 sides, then one of 48 sides, and so on without limit. Now it is obvious that the polygon of 12 sides approaches nearer to a coincidence with the circle, than that of 6 sides. In the same manner the polygon of 24 sides, approaches nearer than that of 12, and the polygon of 48 sides approaches nearer than that of 24, and so on without a limit . But the difference between the first polygon and the circle is a finite or limited quantity, and we have seen that this difference constantly diminishes as we increase the sides. Accordingly if the number of sides were increased to infin, ity, the difference would become nothing; for no one can doubt that the endless diminution of a limited quantity must bring it to nothing. Thus the polygon of an infinite number of sides would not differ from a circle. This idea of a circle agrees with the definition before given of a curved line (10) Damely, that it is made up of infinitely small straight lines. 95. - The perimeters of regular polygons of the same number of sides are to each other as the radii of their circumscribed circles—. By the perimeter of a polygon we mean the sum of its sides. Then we say that the perimter A B C F64 D E F (fig. 64), is to the perimeter G H I K L M as CN is to IO. Suppose the two polygons are hexagons. As they are similar (89) we have BC:HI::CD:I K. Then (66) 6 times B C:6 times HI::CD:I K. But 6 times B C is the perimeter of the first polygon, and 6 times H I is the perimeter of the second. Moreover the triangles B N C and H 0 I are similar. For the angle B NC=HO I since the arcs B C and H I contain the same number of degrees, and B C N=H I O (42) being inscribed in segments containing the same number of degrees. Therefore CD:IK::CN:I 0. Accordingly by making the substitutions in the proportion, 6 times B C:6 times HI::CD:IK, we have the following ; perimeter A B C D E F: perimeter GHIKLM::CN:1 0. As the same reasoning might be employed for any other number of sides thau 6, the proposition is demonstrated. 96. --The circumferences of circles are to each other as their radi:- This follows directly from the two last propositions, for the circumferences of circles are the perimeters of regular polygons of an infinite, and therefore the same number of sides. Moreover the radii of the circumscribed circles become, in this case, the radii of the circles to be compared, the polygons being confounded with the circumscribed circles. Also it may be added thatm-similar arcs are to each other as their radii, By similar arcs we understand those whicb contain the same number of degrees or measure equal angles at the centre. Now from the definition of a degree (15) such arcs are to each other as the circumferences of which they are a part. But these last are to each other as their radii. Therefore similar arcs are to each other as their radii. 39 SECTION SECOND. Surfaces. 97. By the word surface we understand, in the abstract, that magnitude which has length and breadth without thickness. But a more definite idea will be obtained if we introduce motion. Accordingly we say-- surface is the space described by a line moving any other way than lengthwise—. Thus we have the origin of the two dimensions. For the line itself has one dimension, namely, length, and its motion makes another, namely, breadth. Speaking abstractly there is no thickness. But as you cannot make obvious to the senses, that which has absolutely no thickness, it is sufficiently near the truth to say—a surface has length and breadth with an infinitely small thickness. This is analogous to our definition of a line (2), for the infinitely small breadth and thickness of the moving line, would give an infinitely small thickness to the generated surface. Moreover as the boundaries of a line were points, so now, for a similar reason, the boundaries of a surface are lines. 98. There are three kinds of surfaces, corresponding to the three kinds of lines by which we may conceive them to be generated. These are plane, polygonal and curved. We have already defined a plane surface. - That in which any two points being taken, the straight line joining those points, lies wholly in that surface. Thus, if we leave its thickness out of consideration, a sheet of paper perfectly smooth and even, may be taken to represent a plane surface, for in whatever direction we apply the straight edge of a rule to it, the rule will touch it in every point. Such a surface is usually designated by the word plane alone. -A polygonal surface is one which is composed of several planes. If a surface is neither plane nor composed of planes, it is a curved surface. But in order to give a definition which may make a plane the element of all surfaces, we say-a curved surface is one which is composed of infinitely small - planes. This is the point of view in which we shall consider it hereafter. 99. In the first section we considered figures only with reference to the lines and angles of which they consist, and we called those figures equal, which, being applied the one to the other, coincide throughout. In the present section we are to consider figures with reference to the quantity of surface, which they embrace, and we shall call those figures equivalent, which embrace equal quantities of surface. The question then arises, how is this quantity of surface to be estimated ? In other words, how are surfaces measured and compared ? In measuring and comparing lines, (7) we found it necessary to fix upon some quantity of the same kind as a standard of measure, and we called this a linear unit. In like manner if we would measure and compare surfaces, we must fix upon some quantity of the same kind, to serve as a unit of surface or superficial unit. In order to be of the same kind, this unit must have two dimensions, length and breadth; and as that is obviously most simple, in which these two dimensions are the same, geometers have universally adopted, as a superficial unit--a square whose side is a linear unit Accordingly we express the measure of surfaces, by stating the number of square inches, square feet, square yards, &c. which they contain ; meaning thereby the number of squares whose side is an inch, a foot, a yard, &c. The measure of a figure thus expressed, is usually called its 100. The area of a right parallelogram is equal to the product of its base by its altitude - By the altitude of a parallelogram we mean a perpendicular let fall from one side to another parallel side. By the base we mean the side upon which the perpendicular falls. Thus EF F50 (fig. 50) is the altitude, and A D the base. In case of a F65 right parallelogram as A B C D (fig. 65) it follows from the above definition that A B is the altitude and A D the base. Then we say that the area of A B C D=A BXA D. That is A B C D contains as many superficial units as the product of the linear units in A D by those in A B. The superficial unit, as we have just seen, will depend upon the linear unit. Suppose then A D=8 inches and A B=6 inches. Here the superficial unit will be a square inch, as E F G H, and we are to show that it is contained area. 39 26 13 6 in A B C D 6 times 8 or 48 times. If we mark the inches in A D and erect perpendiculars, and do the same with B D, as in the figure, each row_will contain as many squares as there are inches in A D, that is 8; and there will be as many of these rows as there are inches in A B, that is 6. The whole number then is 6 times 8 or 48. The proposition is equally true if A D and A B do not contain an exact number of inches. If for example A D=61 inches and A B=4} inches, still we say that A BC 13 D=6}x4}=28 square inches. For since A D=6) and A B=43 6? we may suppose A D divided into 39 parts each equal to of an inch, and A B divided 6 into 26 equal parts of the same value. Then by erecting perpendiculars as before, we shall have A B C D=39 x26 =1014 squares, the side of which is of an inch. Now of these small squares a square inch contains 6 x6=36, since each row contains 6 squares and there are 6 rows. Accordingly if we divide the whole number 1014 by 36 the number in a square inch, we shall have the number of square inches. Now 1914–28 which was to be proved. Accordingly, since the same reasoning might be employed for any other values of A D and A B, we conclude universally that the area of a right parallelogram is equal to the product of its base by its altitude. If the right parallelogram be a square, since by the definition the base and altitude are the same—we have the area of a square by multiplying one of its sides by itself— Hence the origin of the term square applied to the product of a number multiplied by itself. 101. — The area of any prrallelogram is equal to the product of its base by its altitude. If a parallelogram is not right, it is oblique. Now if we prove that an oblique parallelogram is equivalent to a right parallelogram of the same base and altitude, it will follow that it must have the same measure, namely, its base into its altitude. Accordingly let A B E F (fig. 66) be a right parallelogram, and A B F 66 C D an oblique one, of the same base A B and the same altitude BE. We say they are equivalent Why?-Because the right triangle AF D=B E C (59), since A |